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False. This is an example of integral trace need not imply integrality. Take $K={\mathbb{Q}}_p(\theta )$ with ${\theta }^2=p$, the splitting field of $X^2-p$. Its two conjugates are $\theta ,-\theta$, and its normalized discrete valuation satisfies $v_K(\theta )=1$, $v_K(p)=2$, by the Eisenstein polynomial criterion. Forthe field trace is zero, but $v_K(x)=-1$. Therefore $x$ is not in the valuation ring, despite having integral field trace. Cancellation in a sum of conjugates explains why a field trace test fails.